Answer:
To prove that 3·4ⁿ + 51 is divisible by 3 and 9, we have;
3·4ⁿ is divisible by 3 and 51 is divisible by 3
Where we have;
= 3·4ⁿ + 51
= 3·4ⁿ⁺¹ + 51
-
= 3·4ⁿ⁺¹ + 51 - (3·4ⁿ + 51) = 3·4ⁿ⁺¹ - 3·4ⁿ
-
= 3( 4ⁿ⁺¹ - 4ⁿ) = 3×4ⁿ×(4 - 1) = 9×4ⁿ
∴
-
is divisible by 9
Given that we have for S₀ = 3×4⁰ + 51 = 63 = 9×7
∴ S₀ is divisible by 9
Since
-
is divisible by 9, we have;
-
=
-
is divisible by 9
Therefore
is divisible by 9 and
is divisible by 9 for all positive integers n
Step-by-step explanation:
The answer is B! It passes the vertical line test
68 ( p+11 bc when u add 12 +48 u get 68 and then u times it by 22 it's that what's u get
With what ??? I don’t see nothing
Four different ways to solve a quadratic equation are factoring, completing the square, quadratic formula, and graphing. all methods start with setting the equation equal to zero.
to determine the zeros of a quadratic graph, find the points where the graph of the quadratic equation crosses the x-axis. to determine the vertex, you convert the quadratic form to vertex form, you use this process of completing the square. to determine the axis of symmetry, you cut the graph vertically to see the symmetrical sides.
hope this helps, God bless!